Equal-Area Bisection Problems

Summary: Three chapter-22 problems (II, III, IV) that all ask for a straight cut bisecting a curved region into two equal areas: a chord that bisects its sector (figure 112), an ordinate that bisects a quadrant (figure 113), and a chord through one end of a semicircle that bisects it (figure 114). Problems III and IV reduce to arc-equals-cosine; Problem II yields a fresh constant $s=54^{\circ }18^{\prime }{6}^{\prime \prime }52^{\prime \prime \prime }43^{\prime \prime \prime \prime }33^{\prime \prime \prime \prime \prime }$.

Sources: chapter22, §§532–534. Figures 112 (figures111-114 p. 492), 113 (same), 114 (same).

Last updated: 2026-05-12.

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Problem II — chord bisects sector (§532, figure 112)

Setup. Sector $ACB$ of a unit circle is split by chord $AB$ into the triangle $ACB$ and the segment $AEB$. If $\angle ACB$ is small the triangle dominates; if it is obtuse the segment does. Continuity gives an intermediate angle where the two pieces are equal.

Let $AC=1$, arc $AEB=2s$, so half-arc $AE=BE=s$. Drop the perpendicular $CE$ to the chord. With $F$ on the chord,

$$AF=\sin s,CF=\cos s,\text{triangle }ACB=\sin s\cos s=\frac{1}{2}\sin 2s,$$

while the sector $ACB$ has area $s$ (because radius = 1). Setting sector = $2\times$ triangle:

$$\boxed{s=\sin 2s.}$$

Existence. $\sin 2s$ rises above $s$ near $s=45^{\circ }$ (where $\sin 2s=1>s$) and falls below it as $s\to 90^{\circ }$ (where $\sin 2s=0<s$). Euler notes $s>45^{\circ }$ immediately.

False position. Three passes:

• Degrees: $s\in \{50^{\circ },55^{\circ },54^{\circ }\}$ → between $54^{\circ }$ and $55^{\circ }$; proportion $617584:525041=5^{\prime }:4^{\prime }15^{\prime }$, suggesting $s\approx 54^{\circ }15^{\prime }$.
• Minutes: $s\in \{3257^{\prime },3258^{\prime },3259^{\prime }\}$ (i.e., $54^{\circ }17^{\prime },54^{\circ }18^{\prime },54^{\circ }19^{\prime }$) refines to $s=54^{\circ }18^{\prime }{6}^{\prime \prime }52^{\prime \prime \prime }$.
• Seconds with high-precision logs ($s=54^{\circ }18^{\prime }{0}^{\prime \prime }$ vs $54^{\circ }18^{\prime }10^{\prime \prime }$): $s=54^{\circ }18^{\prime }{6}^{\prime \prime }52^{\prime \prime \prime }43^{\prime \prime \prime \prime }33^{\prime \prime \prime \prime \prime }$.

Final.

$$s=54^{\circ }18^{\prime }{6}^{\prime \prime }52^{\prime \prime \prime }43^{\prime \prime \prime \prime }33^{\prime \prime \prime \prime \prime },2s=108^{\circ }36^{\prime }13^{\prime \prime }45^{\prime \prime \prime }27^{\prime \prime \prime \prime }{6}^{\prime \prime \prime \prime \prime },$$

with $\sin s=AF=BF=0.8121029$, chord $AB=1.6242058$, $\cos s=CF=0.5335143$.

This is the most precise numerical answer in the chapter — six sexagesimal digits past the degree. Modern readers can recognize $108^{\circ }36^{\prime }$ as roughly the chord-of-the-equal-bisecting-cut angle that appears in problems on “lunes” since antiquity.

Problem III — ordinate bisects quadrant (§533, figure 113)

Setup. Quadrant $ACB$ of radius 1. An ordinate $DE\perp AC$ from inside divides it into a region $ADE$ (under arc and below ordinate) and the rest. Find $DE$ that makes the two regions equal.

With arc $AE=s$ and $D=(\cos s,0)$, $E=(\cos s,\sin s)$:

$$\text{area }ADE=\underset{\text{sector }ACE}{\underbrace{\frac{1}{2}s}}-\underset{\text{triangle }CDE}{\underbrace{\frac{1}{2}\sin s\cos s}}.$$

Doubled, set equal to the quadrant $\pi /4$:

$$s-\frac{\pi }{4}=\frac{1}{2}\sin 2s.$$

Reduction to Problem I. Substitute $u=s-45^{\circ }$, then $2s=90^{\circ }+2u$ and $\sin 2s=\cos 2u$:

$$2u=\cos 2u.$$

This is exactly arc-equals-cosine in the doubled variable $2u$, so $2u=s_0=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$, giving

$$u=21^{\circ }10^{\prime }23^{\prime \prime }37^{\prime \prime \prime \prime },AE=s=66^{\circ }10^{\prime }23^{\prime \prime }37^{\prime \prime \prime \prime },$$

with arc $BE=23^{\circ }49^{\prime }36^{\prime \prime }23^{\prime \prime \prime }$ and $CD=0.4039718$, $AD=0.5960281$, $DE=0.9147711$. The half-radius ordinate that bisects a quadrant is $DE\approx 0.915$.

Aside. Repeating the construction on each half quadrant divides the whole circle into eight pieces of equal area (§533 closing).

Problem IV — chord bisects semicircle (§534, figure 114)

Setup. Semicircle $AEDB$ of radius 1; from $A$, draw a chord $AD$ that cuts the semicircle into two equal regions.

With arc $AD=s$ and the radius $CD$ drawn,

$$\text{sector }ACD=\frac{1}{2}s,\text{triangle }ACD=\frac{1}{2}\sin s,$$

so the segment between chord and arc is $AD$-region $=\frac{1}{2}s-\frac{1}{2}\sin s$, equated to half the semicircle $\pi /4$:

$$s-\sin s=\frac{\pi }{2}.$$

Reduction to Problem I. Substitute $u=s-90^{\circ }$, then $\sin s=\cos u$:

$$u=\cos u.$$

So $u=s_0=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$ and $s=132^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$. The angle $ACD=132^{\circ }20^{\prime }47^{\prime \prime }$, complement $BCD=47^{\circ }39^{\prime }12^{\prime \prime }46^{\prime \prime \prime }$, chord $AD=2\sin (s/2)=1.8295422$.

Geometric remark

In all three problems the unknown is an arc, but the equation is a transcendental balance between an arc (linear in $s$, contributed by sector area $s/2$) and a trigonometric quantity (contributed by triangle area $\sin \theta \cos \theta$). This is exactly the kind of equation Euler hopes might collapse into algebraic form if the trig value were lucky — see quadrature-commensurability-remark.

Across the three problems the reduction map is:

Problem	Equation	Substitution	Reduces to
II	$s=\sin 2s$	none (new constant)	—
III	$s-\pi /4=\frac{1}{2}\sin 2s$	$u=s-45^{\circ }$, then double	$2u=\cos 2u$
IV	$s-\sin s=\pi /2$	$u=s-90^{\circ }$	$u=\cos u$

Two of three become Problem I; Problem II requires fresh false-position work because the linear and trigonometric parts don’t separate as cleanly.

Figures

Figures 111–114

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• arc-equals-cosine — the master Problem I that III and IV reduce to.
• false-position-method — the technique used in Problem II.
• quadrature-commensurability-remark — why Euler cares.